Mechanical Advantage of Rack and Pinion

[adityagoti]

Hi All,

Can anyone tell me how to calculate the Mechanical advantage of the rack and pinion system. Thank You.

 

[TheTick]

Either zero or infinite, depending on which side you're driving from.

I don't think you can really say there is a mechanical advantage without considering input force. Otherwise, it is a conversion of torque to/from linear force like a lever arm.

 

[mikekilroy]

T=F*pitch/2*pi

 

[Tmoose]

mechanical advantage starts with motion in versus the motion out.
Friction etc keeps the force in/out ratio being the same.

 

[btrueblood]

"T=F*pitch/2*pi"

No pi in the torque/force equation. Just moment arm, or 1/2 of pitch diameter for the pinion:

Torque = Force * (Pitch dia.) /2

 

[adityagoti]

Thank You all. Looks like there is no MA from rack and pinion. Its only the torque which is based on the diameter of the Pinion. So if this is true, then the linear velocities of both the rack and pinion will always be the same right???

 

[TheTick]

Mechanical advantage is defined by the ratio of like input and output, e.g. torque out/torque in or force out/force in. A system of just one pinion and one rack does have input and output that can be measured this way.

If you are more specific about your system, we can probably get you to a more useful answer.

 

[IRstuff]

But, a rack and pinion is intended to convert rotational force into a linear force, which is then used to produce an angular motion. So the rack and pinion, by itself should not be described in this manner.

However, if you read this, you can determine a figure of merit based on the motion of the wheels relative to the sterring wheel.

http://www.carbibles.com/steering_bible.html

TTFN
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[adityagoti]

Thanks IRstuff, that completely made sense. But the question still remain unanswered of how to determine the Mechanical Advantage, based on the spacing of the teeth.

 

[TheTick]

But the question still remain unanswered of how to determine the Mechanical Advantage, based on the spacing of the teeth.

What color is a 3-sided square?

 

[mikekilroy]

"T=F*pitch/2*pi"

No pi in the torque/force equation. Just moment arm, or 1/2 of pitch diameter for the pinion:

Torque = Force * (Pitch dia.) /2

sorry for not including units to my equation which is correct for his rack/pinion question.....

I was not giving Torque = distance * force

I gave torque= force(#)*pitch (in/rev)/ 2pi (radians/rev).... this converts his thrust force applied to the linear rack into #-in torque on his pinion. there are of course 2*pi radians per revolution so yes pi is included to convert units properly.

 

[mikekilroy]

But the question still remain unanswered of how to determine the Mechanical Advantage, based on the spacing of the teeth.

why not tell us what 'mechanical advantage' means to u? it makes no sense in an engineering question as stated.

but if you are trying to find out something to do with torque vs force relationship and how it changes with 'spacing of teeth,' THAT is easy to answer: none. distance between teeth on rack has ZERO bearing on torque vs. force. NOne. zip, zilch. see my equation in previous post - that shows ALL the relationship there is between these two terms.

 

[drawoh]

Hi All,

Can anyone tell me how to calculate the Mechanical advantage of the rack and pinion system. Thank You.

Compared to what?

You mean a rack and pinion steering system, right?

You rotate steering wheel A, which has gear B. This translates rack C. The rack connects through a linkage, to driving wheel D, some distance from the wheel's pivot point. As an engineer, you should have no problem figuring out how many turns of the steering wheel are required to achieve an angle of rotation of the driving wheel, and what parameters would affect this.

--
JHG

 

[IRstuff]

Again, rotational to linear; there is no mechanical advantage to calculate, and the teeth are irrelevant

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[adityagoti]

Hi All,
Sorry for the confusion. Let me explain clearly what I mean my mechanical advantage.
I need a torque of 100 lb-in at the center of the pinion. I can get an input force of 25lbs from the power source (not more than that) and I can only use 4" dia pinion (limited by space). Therefore I was wondering if there is a way I can choose the a rack and pinion system such that the torque at the center of the pinion will be 100 lb-in. Hope this explation makes my question understandable. Thank you all for the replies.

Regards
Addy

 

[IRstuff]

Then, you need a 4-in pinion. There's not much to do with teeth if they're small enough.

Is this for school?

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[mikekilroy]

you will get your 100#-in at center of 4" pinion with 25# thrust force applied to rack if you move the mechanism to the planet Krypton.

You see, on Krypton, physics has different properties than here. There, T=F*pitch/pi, so T=25#*2*pi*2in/pi=100#

But in order to have their physics work, Krypton rock is deadly to their people and causes the same 100# to act like 5# in our universe.... Nothing is totally free.

 

[drawoh]

adityagoti,

If your pinion must produce a torque of 100lb.in, the torque input through your steering wheel must be 100lb.in. what you ought to be looking for the force to move your rack. For a given force, you can reduce pinion torque by using a smaller pinion. The only other way to reduce pinion torque is to put a gear reducer between your steering wheel and pinion.

Are you sure you know what you are doing? Unless I am horribly misinterpreting your question, this is very basic mechanics of machines.

--
JHG

 

[dinjin]

The torque at the center of the pinion is infinite or indeterminate.
The torque at the pinion gear radius is 2 inch x 25 pounds equals 50 inch-lbs. The torque at 1 inch radius is 100 inch-lbs. And at zero radius, the formula would be 50 inch-lbs times (2 inch radius divided by 0 radius).
T1 = T x R/R1 where T and R are 50 inch-lbs and 2 inch radius respectively.
Help us determine what you are trying to solve. If a motor was rated at 100 inch-lbs, the force on the rack would be 100 inch-lbs divided by the pinion radius. Is this what you are trying to solve?

 

[drawoh]

The torque at the center of the pinion is infinite or indeterminate.
The torque at the pinion gear radius is 2 inch x 25 pounds equals 50 inch-lbs...

According to the OP, the input torque is 100lb.in. The torque output is not affected by radii. The output force is.

--
JHG